# H/wk 13, Solutions to selected problems

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1 H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots. (a) By a direct check we verify that the only roots of x x = 0 in Z 4 are 0 and 1. Thus x x = 0 has roots in Z 4. For every element a of Z we have a = a and hence for every (a, b) Z Z we have (a, b) = (a, b ) = (a, b), so that (a, b) is a root of x x = 0. Thus x x = 0 has Z Z = 4 roots in Z Z. Suppose now that R is an integral domain. It is easy to see that x = 0 and x = 1 are roots of x x = 0 in R. We claim that there are no other roots. Indeed, suppose a R is a root, so that a a = 0 in R. Then 0 = a a = a(a 1). Since R is an integral domain, it follows that either a = 0 or a 1 = 0, that is, either a = 0 or a = 1. Thus x x = 0 has exactly roots in R. By a direct check we verify that x x = 0 has exactly 4 roots in Z 6, namely 0, 1, 3 and 4. (b) Consider the ring R = Z where R = {(a 1, a, a 3,... ) a i Z } and where (a 1, a, a 3,... ) + (b 1, b, b 3,... ) = (a 1 + b 1, a + b, a 3 + b 3,... ) (a 1, a, a 3,... ) (b 1, b, b 3,... ) = (a 1 b 1, a b, a 3 b 3,... ) for a i, b i Z. It is easy to see that R is a commutative ring. Moreover, since for every a Z we have a = a, it follows that for every x = (a 1, a, a 3,... ) R we have is also infinite and commutative, it satisfies all the required proper- Since R = Z ties. x = (a 1, a, a 3,... ) = (a 1, a, a 3,... ) = x and hence x x = 0. Ch. 4.1, Problem 17 In each case factor f(x) into linear factors in F [x]. (a) f(x) = x 4 + 1, F = Z 13. We have 1 = 1 in Z 13. Hence in Z 13 [x] we have x = x 4 1 = x 4 1 = (x 1)(x + 1). Moreover, in Z 13 we have 1 = 5 = 5. Hence in Z 13 [x] we have x = (x 1)(x + 1) = (x 1 )(x 5 ) = (x 1)(x + 1)(x 5)(x + 5). (b) f(x) = x 3 + 1, F = Z 7. In Z 7 we have = 0, so that 1 is a root of f(x) = x in Z 7. Hence (x + 1) f(x) in Z 7 [x]. By performing division with the remainder, we get x = (x + 1)(x x + 1) in Z 7 [x]. We then look for roots of x x + 1 in Z 7. It is easy to see that 3 is such a root since = 7 0 mod 7. Dividing 1

2 x x + 1 with the remainder by x 3 in Z 7 [x] we get x x + 1 = (x 3)(x + ) in Z 7 [x]. Hence x = (x + 1)(x 3)(x + ) in Z 7 [x]. Ch. 4.1, Problem 3 In each case determine the multiplicity of a as a root of f(x). (b) f(x) = x 4 + x + x +, a = 1, R = Z 3. Dividing f(x) by x + 1 with the remainder in Z 3 [x], we get: f(x) = x 4 + x + x + = (x + 1)(x 3 x + ) in Z 3 [x]. Observe that a = 1 is a root of x 3 x + in Z 3 [x]. Dividing x 3 x + by x + 1 in Z 3 [x], we get: x 3 x + = (x + 1)(x x + ) We see that = 1 0 in Z 3, so that a = 1 is not a root of x x + in Z 3 [x]. Hence the multiplicity of a = 1 as a root of f(x) in Z 3 [x] is equal to. Ch. 4.1, Problem 4 If R is a commutative ring, a polynomial f(x) in R[x] is said to annihilate R if f(a) = 0 for every a R. (a) Show that x p x annihilates Z p for a prime p. By Fermat s theorem, for every a Z we have a p a mod p, that is a p = a, a p a = 0 in Z p. Thus indeed x p x annihilates Z p. (b) Show that x 5 x annihilates Z 10. Can be verified via a direct check and also follows from (c). (c) Show that if p is a prime then x p x annihilates Z p. Since p is an odd prime, we have gcd(, p) = 1. Hence by Corollary 1 to Theorem 8 in Ch 3.4 we have that Z p = Z Z p as rings. Thus it suffices to show that x p x annihilates Z Z p. By part (a) we already know that for every b Z p we have b p = b. A direct check shows that for every a Z we have a = a. Therefore for every a Z, b Z p we have (a, b) p = (a p, b p ) = (a, b) and hence (a, b) p (a, b) = (0, 0) so that x p x annihilates Z Z p as required. If p > 3 is a prime, show that x p x annihilates Z 3p. Since p > 3 is a prime, we have gcd(3, p) = 1. Hence by Corollary 1 to Theorem 8 in Ch 3.4 we have that Z 3p = Z3 Z p as rings. Thus it suffices to show that x p x annihilates Z 3 Z p.

3 3 By part (a) we know that x p x annihilates Z p. It is easy to see by a direct check that x p x annihilates Z 3. Indeed, in Z 3 we have 0 p = 0, 1 p = 1 and 1 p = 1, where the last equality holds since p > 3 is a prime and hence p is odd. Therefore for every a Z 3, b Z p we have (a, b) p = (a p, b p ) = (a, b) and hence (a, b) p (a, b) = (0, 0) so that x p x annihilates Z 3 Z p as required. (e) Does x 5 x or x 7 x annihilate Z 35? Since gcd(5, 7) = 1, it follows that Z 35 = Z5 Z 7 as rings. Thus x p x annihilates Z 35 if and only if it annihilates each of Z 5, Z 7. For Z 7 we have 5 = 3 = 4 in Z 7. Thus x 5 x does not annihilate Z 7 and hence it does not annihilate Z 35. Also, for Z 5 we have 7 = 18 = 3 in Z 5. Thus x 7 x does not annihilate Z 5 and hence it does not annihilate Z 35. (f) Show that there exists a polynomial of degree n in Z n [x] that annihilates Z n. Take f(x) = x(x 1)(x )... (x n 1) Z n [x]. Ch. 4., Problem 5 In each case determine whether the polynomial is irreducible over each of the fields Q, R, C, Z, Z 3, Z 5 and Z 7. (a) x 3. We claim that x 3 is irreducible over Q. Since deg(x 3) =, to show that x 3 is irreducible over Q it suffices to prove that x 3 has no rational roots. We have x 3 = (x 3)(x + 3) in R[x]. Since R is an integral domain, it follows that x 3 has exactly two roots in R, namely ± 3. Since both of these roots are irrational, it follows that x 3 has no rational roots and hence it is irreducible over Q. We have x 3 = (x 3)(x+ 3) in R[x] and in C[x]. Hence x 3 is reducible over R and over C. Also, the following holds in Z [x]: and hence x 3 is reducible over Z. Similar, in Z [x] we have: x 3 = x 1 = (x 1)(x + 1) x 3 = x 0 = x = x x and hence x 3 is reducible over Z 3. A direct check shows that x 3 has no roots in Z 5. Indeed, in Z 5 we have 0 3 = 3 0, 1 3 = 0, 3 = 1 0, 3 3 = 1 0 and 4 3 = 3 0. Hence x 3 is irreducible over Z 5. Similarly, a direct check shows that x 3 has no roots in Z 7 and hence it is irreducible over Z 7. (b) x + x + 1

4 4 Via applying the quadratic formula we find the complex roots of x +x+1: x 1, = 1± 3 = 1+± 3i, so that in C[x] we have x +x+1 = (x 1± 3 )(x+ 1± 3 ). Since C is an integral domain, it follows that x 1, = 1± 3 = 1+± 3i are the only roots of x + x + 1 in C. Since none of these roots belong to Q and none of them belong to R, it follows that x + x + 1 has no roots in Q and no roots in R. Since deg(x + x + 1) =, this implies that x + x + 1 is irreducible over Q and it is also irreducible over R. Since x + x + 1 = (x 1± 3 )(x + 1± 3 ) in C[x], it follows that x + x + 1 is reducible over C. A direct check shows that x +x+1 has no roots in Z. Indeed, = 1 0 and = 1 0 in Z. Hence x + x + 1 is irreducible over Z. On the other hand, x + x + 1 has a root in Z 3, namely 1. Hence x + x + 1 is reducible over Z 3. One can also see this directly by checking that in Z 3 [x] we have x + x + 1 = x x + 1 = (x 1). A direct check shows that x + x + 1 has no roots in Z 5. Indeed, in Z 5 we have = 1 0, = 3 0, = 0, = 3 0, = 1 0. Hence x + x + 1 is irreducible over Z 5. On the other hand, x + x + 1 has a root in Z 7, namely. Hence x + x + 1 is reducible over Z 7. (c) x 3 + x + 1. Note that lim x x 3 + x + 1 = and lim x x 3 + x + 1 =. Hence by the Intermediate Value Theorem from calculus there exists x 0 R such that x x = 0. Thus x 3 + x + 1 has a root in R and hence it is reducible over R. Since this real root x 0 also belongs to C, it follows that x 3 + x + 1 is reducible over C. We claim that x 3 + x + 1 is irreducible over Q. Indeed, suppose, on the contrary, that x 3 +x+1 is reducible over Q. Since deg(x 3 +x+1) = 3, it follows that x 3 +x+1 has a root r Q. Then Theorem 9 in Ch 9.1 implies that r = c d where c, d Z and c 1, d 1. Hence c {1, 1} and d {1, 1}. Therefore r = c din{1, 1}. However = 3 0 and ( 1) 3 + ( 1) + 1 = 1 0 in Q, yielding a contradiction. Thus indeed x 3 + x + 1 is irreducible over Q. Ch. 4., Problem 6 Let R be an integral domain and let f(x) R[x] be monic. If f(x) factors properly in R[x], show that it has a proper factorization f(x) = g(x)h(x) where g(x) and h(x) are both monic. Let n = deg(f). Let f(x) = g 1 (x)h 1 (x) be a proper factorization of f(x) in R[x]. Thus degg 1 = m, degh 1 = k where m + k = n and 0 < m, k < n. We have f(x) = x n + a n 1 x n a 0 (since f is monic), g 1 (x) = b m x m + + b 0, h 1 = c k x k + + c 0 where b m, c k R, b m 0, c k 0. Then the leading coefficient of g 1 h 1 is equal to b m c k. Since f is monic, it follows that b m c k = 1 in R. Since R is an integral domain (and thus is commutative) this means that b m and c k are units in R and b m = c 1 k.

5 5 Put g(x) = c k g 1 (x) = c k (b m x m + +b 0 ) = c k b m x m + +c k b 0 = x m + +c k b 0, so that g(x) is monic. Also put h(x) = b m h 1 (x) = b m (c k x k + + c 0 ) = b m c k x k + + b m c 0 = x k + + b m c 0, so that h(x) is monic. We also have f(x) = g 1 (x)h 1 (x) = c k b m g 1 (x)h 1 (x) = c k g 1 (x)b m h 1 (x) = g(x)h(x). Thus we have found a proper factorization of f(x) as a product of two monic polynomials in R[x].

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